Forced Convection

External FlowExternal Flow

Robi Andoyo, STP., M.Sc., Ph.D

Heat Transfer Outside Various Geometries In Forced Convection

• In this cases, a fluid flowing over completely immersed bodies. The heat transfer is occurring between the fluid and the solid only.

e.g: spheres, tubes, and plates.

• The heat transfer flux depends on:

- geometry of the body.

- position of the body. (i.e. front, side, back, etc.)

- flow rate

- fluid properties

• The heat transfer coefficient varies over the body.

Cross Flow on Cylinder

Tube/Immersed bodies

Conditions :

• The long cylinder of diameter D

• Uniform surface temperature Ttemperature Ts

• Cross flow by a free stream fluid of temperature T∞

• Uniform upstream velocity u∞

ReD= ρu∞D/µ

D = Length characteristic, diameter of cylinder

Cross Flow on Cylinder

Tube/Immersed bodies• The average heat transfer coefficient on immersed

bodies is calculated using:

where,

3/1

PrRe NCNNm

Nu =where,

C and m are constant – that depends on various configurations.

NRe – based on undisturbed free fluid velocity.

Fluid properties – evaluated at Tf (film temp.)

Tf = film temperature = (Tw + Tb) / 2.

Hilpert Correlation

Valid for Pr > 0,6

Evaluated at film temperature, Tf

Churchill-Bernstein Correlation

• Covers a wide range of Reynolds and • Covers a wide range of Reynolds and Prandtl numbers

• All properties are evaluated at the film temperature

Cross Flow on Noncircular Tubes

Flow Parallel to Flat Plate

• For laminar region, (i.e. NRe.L < 3 x 105 and NPr > 0.7)

TwTbV

q

31506640 /.NN.

hLN ==

• For turbulent region, (i.e. NRe.L > 3 x 105 and NPr > 0.7)

where L=plate length

31

Pr

50

Re6640 /.

,LNu NN.k

hLN ==

µ

ρνLN L =Re,

31

Pr

80

Re3660 /.

,LNu NN.k

hLN ==

Flow Past Single Sphere

• A single sphere is heated or cooled by a fluid flowing

Tb

past through it.

• Average heat transfer coefficient is predicted using:

where NRe = (1 to 70,000)

NPr = (0.6 to 400)

D= sphere diameter

3/1

Pr

5.0

Re6.00.2 NNk

hDNNu +==

µ

ρνDN =Re

Example 1

• Experiments have been conducted to measurethe convection coefficient on a cylindrical tube12.7 mm in diameter and 94 mm long. Thecylinder is heated internally by an electricalresistance heater and is subjected to a crossresistance heater and is subjected to a crossflow of air in a low-speed wind tunnel. Under aspecific set of operating conditions for which thefree stream air velocity and temperature weremaintained at u∞ 10 m/s and 26.20C,respectively, while the average cylinder surfacetemperature was determined to be Ts 128.40C.

Example 2

Estimate the convective heat transfercoefficient on the outside of oranges(external diameter = 5 cm) whensubmerged in a stream of chilled watersubmerged in a stream of chilled waterpumped around the orange. The velocityof water around an orange is 0.1 m/s.The surface temperature of the orange is

20°C and the bulk water temperature is0°C.

Tube Banks

Tube Banks

Tube Banks

2

D S )

2

S( S S T

5.0

2T2

LD

+<

+=

Tube Banks

ST/D=SL/D=2,0ST/D=SL/D=1,25 ST/D=SL/D=1,5

n n n

Example 3

• Air at 1 atm and 50C flow across a bank of tubes

15 rows high and 5 rows deep at a velocity of 7

m/s measured at a point in the flow before the

air enters the tube bank. The surfaces of the

tubes are maintained at 500C. The diameter oftubes are maintained at 500C. The diameter of

the tubes is 2.54 cm they are arranged in a in

line manner so that the spacing in both and the

normal and pararel direction to the flow is 3.81

cm. calculate the total heat transfer per unit

length for the tube bank.

Free/Natural Convection

Basic Principles

• No forced motion

• Induced by buoyancy forces, which arise from density differences caused by temperature variations in the fluidtemperature variations in the fluid

• Flow velocities are generally much smaller than those associated with forced convection

• Heat transfer rates are smaller than forced convection

Natural Convection Heat Transfer

• Occurs when there is a temperature different between

fluid and solid surface.

• Density differences in the fluid arising from the heating

process provide the buoyancy force required to move the

fluid.fluid.

• The dimensionless number, Grashof number (NGr) has

been introduced to obtain the heat transfer coefficient, h

for natural convection.

The boundary layer

• Heated vertical plate that is immersed ina cooler fluid (Ts > T∞ )

• The fluid density gradient and thegravitational field create the buoyancyforce that induces the free convectionboundary layer flow in which the heatedfluid risesfluid rises

• The velocity at the surface is zero (the no-slip condition)

• With increasing distance y from the plate, the velocity increases to a maximum value, then decreases to zero (no-shear condition) the boundary layer thickness (δ).

The boundary layer

• The hydrodynamic and thermalboundary layer flows are coupled

• Thermal effects induce flow, which inturn affects the temperaturedistribution.

• At y = 0, the fluid is at the surfacetemperature, T , and the profile has atemperature, Ts, and the profile has agradient at the surface (y = 0), whichdecreases in the y direction as thetemperature eventually decreases tothat of the quiescent fluid, T∞.

• The temperature becomes zero, noheat transfer.

The Grashof number, NGr

• NGr is the ratio between the bouyancy forces and viscous forces

• NGr in natural convection is analogous to the Reynolds number in forced convection

• The key buoyancy-related parameters : the • The key buoyancy-related parameters : the temperature difference (Ts-T∞) and the volumetric thermal expansion coefficient (β), for ideal gas ρ = p/RT

β = 1/ρ x p/RT2 = 1/T

• β is a thermodynamic property relating the variation of density with temperature

• Grashof number, NGr;

NGr =L

3ρ2gβ∆T

µ2

Grashof Number for planes and

cylinders

Where,ρ = fluid density

fT

1=β

ρ = fluid densityµ = fluid viscosityL = length∆T = temperature difference between

Tw and Tb

g = 9.80665 m/s2

β = volumetric coefficient of expansion

Predicting h

NNu = h dc / k = a (NRa)m

Where a and m are constants; NRa is the Rayleigh number

NRa = NGr x NPr

The Grashof number, NGr (ratio between the bouyancy forces and viscous forces) is defined as :

Grviscous forces) is defined as :

NGr = dc3 ρ2 g β ∆T / µ2 = dc

3 g β ∆T / ṽ2

NPr = µ Cp/k

dc is characteristic dimension (m); ρ is density (kg/m3); g is acceleration due to gravity (9,8 m/s2); β is coefficient of volumetric expansion (K-1); ∆T is the temperature difference between wall and the surrounding bulk/T∞ (0C); ṽ is kinematic viscosity (m2/s) and µ is dynamic viscosity (Pa.s)

All properties are evaluated at film temperature, Tf

The Vertical Plate

General equation

The Churchill-Chu correlation may be applied over the entire range of RaL and has the form

This correlation can be applied for vertical cylinder when

D ≥ 35L/Gr0,25

The Horizontal Plates

• If the plate is horizontal, the buoyancy force is

normal to the surface

• The flow patterns and heat transfer rate depend

strongly on whether the surface is hot or cold strongly on whether the surface is hot or cold

and on whether it is facing upward or downward

• Buoyancy force is normal, instead of parallel, to the plate.

• Flow and heat transfer depend on whether the plate is heated or cooled and

whether it is facing upward or downward.

• Heated Surface Facing Upward or Cooled Surface Facing Downward

The Horizontal Plates

sT T∞> sT T∞<

( )1/ 4 4 70.54 10 10L L LNu Ra Ra= < <

( )1/ 3 7 110.15 10 10L L LNu Ra Ra= < <

• Heated Surface Facing Downward or Cooled Surface Facing Upward

The Horizontal Plates

sT T∞> sT T∞<

( )1/ 4 5 100.27 10 10L L LNu Ra Ra= < <

L = A/P (A = Surface area, P = Perimeter)

Hot Surface Facing Downward or Cold Surface

Facing Upward (Cases A and B)

Hot Surface Facing Upward or Cold Surface

Facing Downward (Cases C and D)

The Horizontal Cylinder

• For a heated cylinder, the boundary

layer development begins at θ = 0 and

concludes at θ < 180

• For the long, horizontal cylinder :

• The Churchill-Chu correlation is

recommended for a wide Rayleigh

number range

The Sphere

• Boundary layer development for the isothermal

sphere is similar to that for the cylinder with the

formation of a plume

• The Churchill correlation is recommended for • The Churchill correlation is recommended for

estimating the average convection coefficient :

Example 1

Estimate the HT coefficient for convectiveheat loss from a horizontal 10 cm diameterpipe. The surface temperature of theuninsulated pipe is 1300C and the airuninsulated pipe is 1300C and the airtemperature is 300C

Example 2

Vertical wall of a peanut drier is made from metal and heated, used to heated air chamber in another side, has a height of 0.71 m and a width of 1.02 m and reaches 0.71 m and a width of 1.02 m and reaches a temperature of 2320C. If the room temperature is 230C, estimate the convection heat rate from the wall to the air chamber.

Example 3

A horizontal, high-pressure steam pipe of 0.1 m outside diameter passes through a large room whose wall and air temperatures are 230C. The pipe has an temperatures are 230C. The pipe has an outside surface temperature of 1650C. Estimate the heat transfer from the pipe per unit length.

Example 4

A home heater consists of a vertical plate with dimensions 0,5 m x 1,0 m. If the temperature on the surface of the plate is maintained at 70 0C and the room maintained at 70 0C and the room temperature is equal to 21 0C, calculate the heat transferred to the room.

Estimation of Overall HT

Coefficient

• A thermal resistance (Rt) term for convective HT may defined in a similar manner as in conductive HTmanner as in conductive HT

q = h A (Ts - T∞) = (Ts - T∞) / (1/hA)

(1/hA) = (Rt)convection

Estimation of Overall HT

Coefficient

• Convective and

conductive HT may occur

simultaneously

q = U1 A1 (T1 – T2)

U1 = overall HT coefficient based on the inside area (A1)of the pipe

q = (T1 – T2) / (1/U1A1)

• 1/U1A1 = overall thermal resistance 1/U1A1 = + +

Example 5

• A 2,5 cm inside diameter pipe is being used to convey a liquid food at 800C. The inside convective HT coefficient is 10 W/(m2 .0C). The pipe (0,5 cm thick) is made of steel (k = 43 W/(m.0C)). The made of steel (k = 43 W/(m.0C)). The outside ambient temperature is 200C. The outside convective HT coefficient is 100 W/(m2 .0C). Calculate the overall HT coefficient and the heat loss from 1 m length of the pipe.

Thank you!